Li for a cylindrical shell REVIEWED

This is the Title PhL 3.26.05 This is now installed in Appendix C, and I found external verification for the result in a 1922 book! A new problem: Consider a cylindrical shell conductor of inner radius a and thickness δ. What is the DC internal inductance of such a shell? First, compute the H field. The total area of the shell is this A = π(a+δ)2 - πa2 = π [(a+δ)2 - a2] If we assume a constant current density in this shell, we find that Jz(r) = I/A = The current enclosed by a loop of radius r is given by Ienc(r) = ∫ dS Jz = 2π!Syntax Error, Irdr Jz(r)dS = 2π !Syntax Error, Ir' dr' = 2π [r2-a2] (1/2) = I The current enclosed at radii ranging from a to a+δ is this Ienc(r) = I Ienc(r) = I which I could have started off with, it is a completely obvious fact. Now compute Hθ(r) 2πrH(r) = current enclosed = Ienc(r) = I dim OK H(r) = (I/2πr) u = (1/2)μ1 H2 = (1/2) μ1 ()2 (1/2π)2 I2 U = ∫dS u = 2π !Syntax Error, Idr r u = (1/2) μ1 ()2 (1/2π) I2 !Syntax Error, Idr r Maple does the integral which result we re-express in terms of the ratio b ≡ δ/a, Given this expression for J, we have U = (1/2) μ1 ()2 (1/2π) I2 J = (1/2) μ1a-4 ()2 (1/2π) I2 J = (1/2) μ1a-4 () (1/2π) I2 J and then U = (1/2)L I2 so L = μ1a-4 () (1/2π) J(b) In the limit a→0 we have b→∞ and then J(b) = (b4a4)/4 so L = μ1a-4 b-4 (1/2π) (b4a4)/4 = μ1 (1/8π) which is the correct result for a solid round wire. For a thin cylindrical shell, we instead take b→ 0. Maple tells us that so that in this limit we get Li = μ1a-4 (1/2π) { J(b) } = μ1a-4 (1/2π) (a4b/3) = μ1 (1/2π) (b/3) = (μ1/6π)b = (μ1/6π)(δ/a) = (μ1/8π)(4/3)(δ/a) Now if I crudely think of this as a rolled up microstrip of width w = 2πa then I am getting Li = (μ1/8π) (4/3) (2πδ/w) = (μ1/8π) [8π/3] (δ/w) This is not too different from the quoted result for an unrolled microstrip, Li = (μi/8π) [(4π/3) a/b ] a/b << 1 (flat wire) // ≈ (μi/8π) [ 4.2 a/b ] (C.4.11) I think my result is the right one! It is true that the flat wire and the annulus have different H field distributions so that might account for the difference. Amazingly I found as PDF which has these quotes My result was μ1(1/4π)[ 3-4ln2] (δ/a), but their result is μ1(1/4π) (δ/a). They give this reference for their results and I think I have access to that stuff now. I download the thing and save it. I then do OCR with Acrobat so I can search the thing. None of them come searchable for some reason. It is 48 pages long. But I don't see this formula in there. The result quoted does match my Appendix K high frequency result Li(ω) = (1/ω) Im(Zs) = (1/ω) = μ1σ1δ2/2 = μ1 (1/4π) (δ1/a1) except the quote shows some R' thing. I cannot find anything like this in that PDF! But I know exactly what they are saying. So no need for The BSTJ thing. There is no H field inside my cylinder by the way. I think my result is correct and that airline PDF paper is a red herring result that is wrong. So I will redo each step in the a = 0 limit and see where it disagrees with the round wire step.